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RMO 2012 solution to Question No. 6

6. Find all positive integers n such that \((3^{2n} + 3 n^2 + 7 )\) is a perfect square.
Solution:
We use the fact that between square of two consecutive numbers there exist no perfect square. That is between \((k^2 )\) and \(((k+1)^2 )\) there is no square.
Note that \((3^{2n} = (9^n)^2 )\) and  \(((9^n + 1)^2 )\) are two consecutive perfect square and \((3^{2n} + 3 n^2 + 7 )\) is always a number between them for n > 2 (easily proved by induction).

Hence the only solution is n = 2.

December 4, 2012

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